如何在 TypeScript 中声明“任何”模块？

How to declare "any" module in TypeScript?

提问by user1338054

I need to migrate step by step some large project from js to typeScript.

我需要将一些大型项目从 js 逐步迁移到 typeScript。

I rewrite on of the files in ts and i want to specify that other files at this moment can contain any content.

我重写了 ts 中的文件，我想指定此时其他文件可以包含任何内容。

For example something like that:

例如这样的事情：

declare module jsModule:any;
var obj:jsModule.cls = new jsModule.cls()

But it does not work in this moment. I need to specify each exported class/function/variable in module declaration.

但它在这一刻不起作用。我需要在模块声明中指定每个导出的类/函数/变量。

Can i declare external module as "any" in some fast way?

我可以以某种快速的方式将外部模块声明为“任何”吗？

回答by Ryan Cavanaugh

For an external module with no exposed types and any values:

对于没有公开类型和任何值的外部模块：

declare module 'Foo' {
  var x: any;
  export = x;
}

This won't let you write foo.cls, though.

但是，这不会让你写foo.cls。

If you're stubbing out individual classes, you can write:

如果您要剔除单个类，则可以编写：

declare module 'Foo' {
    // The type side
    export type cls = any;
    // The value side
    export var cls: any;
}